---
title: "Continuous flushing settling basin, 95% efficiency"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:settling-basin-continuous-60cumec-velikanov"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-continuous-60cumec-velikanov
site: BCT Engineering
---

# Continuous flushing settling basin, 95% efficiency

Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants

- Asked in: 2070 Chaitra, 6+3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-continuous-60cumec-velikanov
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

Find out the dimension of a continuous flushing settling basin for a high head project in a Himalayan river which utilizes a discharge of 60 m³/s and head of 300 m; the sediment particles larger than 0.15 mm have to be trapped with efficiency 95% in the basin. Consider the effect of turbulence and check the length of the basin using Velikanov's relation for the density of the silty water of 1.105 ton/m³. Draw plan and section of the basin showing major components.

## Short answer

With 60 m³/s plus 10% continuous flushing (66 m³/s), trapping >0.15 mm at 95% efficiency, v=0.17 m/s, depth 6 m. Three chambers each 22 m wide; turbulence length 176 m, Velikanov with buoyancy 223 m, adopt 225 m. Dimensions: 3 chambers 22 m × 225 m × 6 m.

## Answer

**Given**: design discharge $Q=60.0$ m³/s and gross head 300 m; particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (assumed for 0.15 mm sand, as in similar designs). The basin has continuous flushing, so 10% extra flow (6 m³/s) is added. The required trap efficiency is 95% for particles larger than 0.15 mm, and the effect of turbulence is included. The length is checked by Velikanov's relation with $\lambda=1.5$ (assumed) and silty water of density 1.105 t/m³.

Flushing water of 10% is added for continuous flushing, so the design flow of the basin is $Q'=60.0+6.0=66.0$ m³/s.

#### Step 1: Horizontal velocity (Camp)
For $0.1<d<1$ mm, $v=a\sqrt d$ with $a=44$ (v in cm/s, d in mm):
$$
v=44\sqrt{0.15}=17.04\ \text{cm/s}=0.1704\ \text{m/s}
$$
This is within the usual 0.1-0.4 m/s range, so deposited sediment is not re-suspended.

#### Step 2: Cross-section and width
Assume a flow depth $H=6.0$ m in the settling zone (sediment storage is extra).
$$
B=\frac{Q'}{vH}=\frac{66.0}{0.1704\times6.0}=64.55\ \text{m}
$$
Provide $n=3$ chambers, so that one can be taken out for inspection while the others work: $b=B/n=21.52$ m each, adopt 22.0 m.

#### Step 3: Length without turbulence
$$
L_0=\frac{vH}{w}=\frac{0.1704\times6.0}{0.0150}=68.2\ \text{m}
$$

#### Step 4: Effect of turbulence
Turbulence reduces the effective fall velocity: $w'=w-\alpha$, with $\alpha=0.132\,v/\sqrt{H}$.
$$
\begin{aligned}
\alpha&=\frac{0.132\times0.1704}{\sqrt{6.0}}=0.0092\ \text{m/s}\\
w'&=0.0150-0.0092=0.0058\ \text{m/s}\\
L&=\frac{vH}{w'}=\frac{0.1704\times6.0}{0.0058}=175.8\ \text{m}
\end{aligned}
$$

#### Step 5: Check by Velikanov's method
The silty water has a density of 1.105 t/m³ (sediment 2.65 t/m³). This reduces the effective fall velocity by the buoyancy ratio $\frac{2.65-1.105}{2.65-1}=0.936$, so $w_s=1.405$ cm/s.
$$
L_V=\frac{\lambda^2v^2(\sqrt H-0.2)^2}{7.51\,w_s^2}=\frac{1.5^2\times0.1704^2\times(\sqrt{6.0}-0.2)^2}{7.51\times0.01405^2}=223.2\ \text{m}
$$

The Velikanov length (223.2 m) is 27% greater than the turbulence length (175.8 m). The larger of the two is adopted and rounded up to 225 m.

#### Adopted dimensions
| Item | Value |
|---|---|
| Flow velocity $v$ | 0.170 m/s |
| Settling depth $H$ | 6.0 m (plus about 1.0 m for the flushing hoppers) |
| Number of chambers | 3 |
| Width of each chamber | 22.0 m |
| Length of settling zone | 225 m |
| Length/width ratio | 10.2 |
| Inlet transition | gradual, with guide vanes (side slope not steeper than about 1:5) |

Total volume of water in the basin $\approx 89100$ m³.

```text
PLAN (continuous flushing outlets along each chamber)
          inlet        settling chambers      outlet
 canal   transition  +=================+   transition
 =====>  \         / |  chamber 1      | \         /===> to
         /  guide  \ +-----------------+ /  weir   \    headrace
 =====>  \  vanes  / |  chamber 2      | \         /
         /         \ +-----------------+ /         \
                     |  chamber 3      |
                     +=================+
  trash-                  |  |  |  flushing conduits (2 m3/s each)
  rack                    v  v  v  --> flushing channel --> river
```

```text
LONGITUDINAL SECTION
 inlet          settling zone (depth H)      outlet
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  WL
 \    --->   --->   --->   --->   --->   --->  |
  \  ______________________________________   | weir
   \/ \/ \/ \/ \/ \/ hoppers with bottom outlets \__|__ to headrace
   ==== flushing conduit (continuous flow) ====> river
```

**Flushing arrangement**: the floor of each chamber is formed into longitudinal hoppers with a slot or a row of openings at the bottom leading to a flushing conduit below. About 2 m³/s per chamber (6 m³/s in total) is drawn off continuously through these openings and carries the settled sediment back to the river, so the chambers do not have to be taken out of service for flushing. A gate on each flushing conduit controls the flushing flow, and inlet gates allow one chamber to be isolated and emptied for inspection while the other two keep supplying the plant.

The adopted length is greater than the Velikanov length, so more than 95% of the particles coarser than 0.15 mm are expected to be trapped.

**Answer**: 3 chambers, each about 22.0 m wide x 225 m long x 6.0 m deep (settling zone), with $v=0.170$ m/s and 6 m³/s of continuous flushing water.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What horizontal velocity is used for d = 0.15 mm (Camp: v = a sqrt(d), a = 44, v in cm/s, d in mm)?
   A. v = 44 sqrt(0.15) = 17.04 cm/s = 0.1704 m/s
   B. v = 44 x 0.15 = 6.6 cm/s = 0.066 m/s
   C. v = 44 sqrt(0.15) = 17.04 m/s
   D. v = 44/sqrt(0.15) = 114 cm/s = 1.14 m/s
   Answer: A. For 0.1 < d < 1 mm Camp gives v = a sqrt(d); the result must be converted from cm/s to m/s and lies in the 0.1-0.4 m/s range so deposits are not resuspended. (Key point: Camp horizontal velocity)

2. For a continuous flushing basin with Q = 60 m^3/s, what flow is used to find the width, and what is B for H = 6 m?
   A. Q = 60 m^3/s only, B = 58.68 m
   B. Q' = 60 - 6 = 54 m^3/s, B = 52.81 m
   C. Q' = 66 m^3/s, B = 66/0.1704 = 387 m
   D. Q' = 60 + 10% = 66 m^3/s, so B = 66/(0.1704 x 6) = 64.55 m
   Answer: D. Continuous flushing needs extra water (10%) that also passes through the basin, so the design flow is increased to 66 m^3/s. (Key point: Design flow with flushing)

3. What is the effective fall velocity w' with turbulence and the resulting length?
   A. w' = w + alpha = 0.0242 m/s, L = 42.3 m
   B. w' = alpha = 0.0092 m/s, L = 111.1 m
   C. alpha = 0.132 v/sqrt(H) = 0.0092 m/s, w' = 0.0058 m/s, L = vH/w' = 176.3 m
   D. w' = w = 0.0150 m/s, L = 68.2 m
   Answer: C. Turbulence is subtracted from the fall velocity (w' = w - alpha) and the length is vH/w'. (Key point: Turbulence correction)

4. How is the fall velocity corrected for silty water of density 1.105 t/m^3, and what does the Velikanov check give?
   A. w_s = 1.5 x (2.65 - 1.105)/(2.65 - 1) = 1.405 cm/s, giving L_V = 223.0 m, so 225 m is adopted
   B. w_s = 1.5 x 1.105 = 1.66 cm/s, giving a shorter basin than turbulence
   C. w_s = 1.5 cm/s unchanged; density has no effect
   D. w_s = 1.5 x (2.65 - 1.105) = 2.32 cm/s, giving L_V below 100 m
   Answer: A. Dense silty water gives less buoyancy difference, so fall velocity drops to 1.405 cm/s and the Velikanov length (about 223 m) exceeds the turbulence length (175.8 m). (Key point: Velikanov with silty water)

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